A committee of 5 members is to be formed from 4 ladies and 6 gentlemen what is the probability

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Permutation
Combination
Permutation and Combination

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Hint: Here, using the given condition(at least one lady should be there) we will have four possible ways to form a committee with at least one lady. After that we will add them. Doing this will get the answer.Complete step-by-step answer:According to the question we have to make a committee of 5 and in each committee formed there must be at least one lady. There are 6 gentlemen and 4 ladies.We can select 5 members for the committee and include at least one lady in the following four ways as done below:(1) 1 lady and 4 gentlemen(2) 2 ladies and 3 gentlemen(3) 3 ladies and 2 gentlemen(4) 4 ladies and 1 gentlemenSo, for the first condition we have to select 1 lady from 4 ladies and 4 gentlemen, therefore we can do:$^4{C_1}{ \times ^6}{C_4} = \dfrac{{4!}}{{1!(3!)}} \times \dfrac{{6!}}{{4!(2!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 4 \times 3 \times 2 \times 1}} = 60$ ….(1)For the second condition we have to select 2 ladies from 4 ladies and 3 gentlemen, therefore we can do:$^4{C_2}{ \times ^6}{C_3} = \dfrac{{4!}}{{2!(2!)}} \times \dfrac{{6!}}{{3!(3!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 2}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 3 \times 2 \times 1}} = 120$ ……(2)For the third condition we have to select 3 ladies from 4 ladies and 2 gentlemen, therefore we can do:$^4{C_3}{ \times ^6}{C_2} = \dfrac{{4!}}{{3!(1!)}} \times \dfrac{{6!}}{{4!(2!)}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}} = 60$……(3)For the fourth condition we have to select all 4 ladies from 4 ladies and 1 gentleman, therefore we can do:$^4{C_4}{ \times ^6}{C_1} = \dfrac{{4!}}{{4!}} \times \dfrac{{6!}}{{5!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = 6$ ……(4)On adding (1), (2), (3) & (4) we get the total number of ways of forming committee are:60 + 120 + 60 + 6 = 246 Hence, the required number of committees is 246.Hence the correct option is B.Note: Whenever you face such types of problems you have to think of the number of ways then according to the condition provided you have to select people with the help of combinations as it is used for selections. This way will take you to the right answer.

Answer : Since the committee of 5 is to be formed from 6 gents and 4 ladies.

(i) Forming a committee with at least 2 ladies Here the possibilities are

2 ladies and 3 gents
3 ladies and 2 gents
4 ladies and 1 gent

The number of ways they can be selected

= 186 ways= 4C2 X6C3 + 4C3X6C2 + 4C4X6C1

(ii) The number of ways in this case is

0 ladies and 5 gents
1 lady and 4 gents
2 ladies and 3

The total ways are

= 4C0 X6C5 + 4C1X6C4 + 4C2 X 6C3

= 186 ways.

Answer: (2) 246

Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination.

Permutation

In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. A permutation is the choice of r things from a set of n things without replacement and where the order matters.

nPr = (n!) / (n-r)!